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slega
2 months ago
15

The indicated function y1(x) is a solution of the given differential equation. Use reduction of order or formula (5) in Section

4.2, y2 = y1(x) e−∫P(x) dx y 2 1 (x) dx (5) as instructed, to find a second solution y2(x). y'' − 8y' + 16y = 0; y1 = e4x
Mathematics
1 answer:
AnnZ [12.3K]2 months ago
5 0
y2 = C1xe^(4x) Step-by-step explanation: Knowing that y1 = e^(4x) satisfies the differential equation y'' - 8y' + 16y = 0, we need to derive the second solution y2 using the reduction of order technique. Let y2 = uy1. Since y2 is a solution to the differential equation, it holds that y2'' - 8y2' + 16y2 = 0. By substituting for y2, its derivatives become y2 = ue^(4x), y2' = u'e^(4x) + 4ue^(4x), and y2'' = u''e^(4x) + 8u'e^(4x) + 16ue^(4x). Plugging these into the differential equation gives us u''e^(4x) = 0. Let w = u', so w' = u''. This results in w' e^(4x) = 0, leading to w' = 0. Integrating gives w = C1. Since w = u', this implies u' = C1, and integrating once more results in u = C1x. Therefore, y2 = ue^(4x) becomes y2 = C1xe^(4x), which is the second solution.
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